describe exponential and logistic population growth, distinguish density-dependent from density-independent limiting factors, and ask testable questions about what limits the tule elk herd at Point Reyes.
Instruction
The phenomenon. Tule elk live only in California. Before the Gold Rush, hundreds of thousands grazed the Central Valley and coastal hills; by the 1870s market hunting and ranching had reduced them to a tiny remnant, famously protected on a Kern County cattle ranch. Their recovery is a conservation success, and one chapter of it is visible at Point Reyes National Seashore. In 1978 the National Park Service released ten tule elk onto Tomales Point, a fenced peninsula. With plenty of grass and no hunting, the herd grew for three decades, to roughly 540 animals by 2012 according to park counts. Then came the 2012-2014 drought, and counts fell to under 300 by 2014, a loss of nearly half. Our question for the unit: what sets the size of a population, and why did this one crash?
Exponential growth. When resources are plentiful, each individual adds offspring at a roughly constant rate, and the population grows faster the bigger it gets. The model is N = N0ert, where r is the per-capita growth rate. If the herd had grown exponentially from 10 in 1978 to 540 in 2012, then r = ln(540/10) ÷ 34 ≈ 0.117 per year, which means the herd would double about every ln 2 ÷ 0.117 ≈ 5.9 years. No population can keep that up for long.
Logistic growth and carrying capacity. As a population grows, each individual gets less food, water and space. Growth slows and levels off near the carrying capacity, K, the largest population the environment can support over time. The logistic model captures this: the growth rate is rN(1 − N/K). With r = 0.4 and K = 500, a population of 100 grows by 0.4 × 100 × (1 − 0.2) = 32 a year; at 250 it grows by 50, the fastest; at 450 it grows by only 18. The graph is an S-shaped curve.
What limits a population? Density-dependent factors get stronger as the population gets denser: competition for grass, the spread of disease, parasites. Density-independent factors affect a population whatever its size: a drought, a fire, a freeze. The two interact. A drought lowers the grass supply, which lowers the carrying capacity; a herd that fit comfortably at its old K is suddenly far above the new one. In the Tomales Point case, park staff pointed to drought and the limited availability of fresh water and forage inside the fence. You will weigh that evidence yourself.
Scale matters. HS-LS2-1 asks about carrying capacity at different scales: a cup of duckweed, a fenced peninsula, all of California. The same mathematics applies, but the limiting factors differ.
Asking questions. Why did the elk die? is a start. Is the number of elk in a year related to the rainfall in the previous winter? is testable with park counts and weather records.
Formative check
Work through these before moving on. They are not graded — they tell you, and your teacher, whether the standard below has landed yet.
A herd near its carrying capacity faces a two-year drought that halves the grass supply. What happens?
Which is a density-dependent limiting factor?
With r = 0.4 per year and K = 500, a population of 250 grows by individuals per year.
Sketch the growth curve you think the Tomales Point herd followed from 1978 to 2014, labeling the exponential phase, the approach to carrying capacity and the drought. Then write two testable questions your sketch raises.
How confident are you that you can describe exponential and logistic growth and explain how limiting factors set a carrying capacity?
Practice
Work these on paper or in your notebook, then open Check your answer. Aim for all of Fluency and Application; try at least one Challenge.
Printable version: this unit’s practice workbook (PDF)
Fluency
Build speed and accuracy with the core skill.
- A population of 1,000 has 39 births and 56 deaths in a year (no migration). Find the growth rate per individual and the population after one year.
Check your answer
Answer: r = −0.0170 per year; 983r = (births − deaths)/N. - A population of 2,000 has 51 births and 57 deaths in a year (no migration). Find the growth rate per individual and the population after one year.
Check your answer
Answer: r = −0.0030 per year; 1,994r = (births − deaths)/N. - A population of 50 grows exponentially with a per-capita rate of 0.05 per year (continuous). Estimate its size after 6 years.
Check your answer
Answer: ≈ 67N = N0ert. - Classify each limiting factor as density-dependent or density-independent: (a) a disease that spreads between elk; (b) a severe frost; (c) competition for water at a single pond; (d) a wildfire.
Check your answer
Answer: (a) density-dependent; (b) density-independent; (c) density-dependent; (d) density-independentDensity-dependent effects grow stronger as more individuals crowd together.
Application
Use the skill in context. Show your reasoning.
- Ten tule elk were released at Tomales Point in 1978, and park counts reached roughly 540 in 2012. Assuming exponential growth, find the per-capita growth rate r and the doubling time.
Check your answer
Answer: r ≈ 0.117 per year; doubling time ≈ 5.9 yearsr = ln(540/10) ÷ 34 = 3.989 ÷ 34 = 0.1173. Doubling time = 0.693 ÷ 0.1173 = 5.9 years. - Using the logistic model with r = 0.2 per year and K = 540, find the yearly growth when the herd is 100, 270 and 500 elk.
Check your answer
Answer: About 16, 27 and 7 elk per yearrN(1 − N/K): 0.2 × 100 × (440/540) = 16.3; 0.2 × 270 × 0.5 = 27; 0.2 × 500 × (40/540) = 7.4. - The herd fell from about 540 in 2012 to 286 in 2014 (park counts). What percentage of the herd was lost?
Check your answer
Answer: About 47%(540 − 286) ÷ 540 = 254/540 = 0.470.
Challenge
Stretch problems. Expect to think before you write.
- If the herd had kept growing exponentially at r = 0.117 per year from 540 in 2012, how many elk would there have been 10 years later? Use your answer to explain why exponential growth cannot continue on a fenced peninsula.
Check your answer
Answer: About 1,750 elk. The same grassland could not feed more than three times as many animals; food, water and space would limit growth, so the curve must level off near a carrying capacity or crash.540 × e0.117 × 10 = 540 × e1.17 = 540 × 3.22 ≈ 1,745.
Review
Keep earlier skills sharp.
- In pea plants, where purple flowers (A) are dominant to white (a), and in a trait where A (dominant) masks a (recessive), cross AA × Aa. Give the genotype ratio and the probability of the dominant phenotype.
Check your answer
Answer: AA: 2/4, Aa: 2/4; P(dominant phenotype) = 1Draw the 2 × 2 Punnett square.